Critical points/ Abs. max/min values

Hey guys so I need help on two questions that I'm sort of stuck on.

1. Find the critical numbers of the given function: f(x) = 2x^{3}+x^{2}+2x

So I got f'(x) = 6x^{2}+2x+2

Factored out the 2(3x^{2}+x+1)

I can't factor this out anymore, so is the answer there are no critical numbers?

2. Find the absolute max and min values of f on the given interval.

√ = square root

f(t) = ^{3}√t * (8-t), [0,8]

So I used product rule on this one and got

f'(t) = 3/2√t * (8-t) - ^{3}√t = 0

3(8-t)-3√t = 0

So, I'm uncertain about the critical numbers here as well are they 8 and 0?

Re: Critical points/ Abs. max/min values

1) If you sketch a cubic equation you will see that it has no maximum or minimum.

2) There's a mistake in your differentiation of t^{1/3}

Re: Critical points/ Abs. max/min values

Quote:

Originally Posted by

**Shakarri** 1) If you sketch a cubic equation you will see that it has no maximum or minimum.

2) There's a mistake in your differentiation of t^{1/3}

Yeah I just noticed the derivative was wrong for number 2.

For 1 do I just leave my answer as it is? and write no critical points then?

Re: Critical points/ Abs. max/min values

For the first one the reason is that the quadratic 3x^2+x+1 does not have real roots. Its discriminant is negative so it has imaginary roots and hence no critical points.

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Re: Critical points/ Abs. max/min values

Re: Critical points/ Abs. max/min values

Quote:

Originally Posted by

**ibdutt**

Quote:

Originally Posted by

**ibdutt** For the first one the reason is that the quadratic 3x^2+x+1 does not have real roots. Its discriminant is negative so it has imaginary roots and hence no critical points.

Have I told you how much I love you?

Haha, thanks a lot for both of the questions. You are the best!

Re: Critical points/ Abs. max/min values

So the only critical point for question 2 will be 0, and when looking for absolute max/ min, with the intervals, the f(x) would give 0 for both the numbers in the interval. So, is it possible to have a abs. max/min on the same point?

Re: Critical points/ Abs. max/min values

Quote:

Originally Posted by

**Oldspice1212** So the only critical point for question 2 will be 0, and when looking for absolute max/ min, with the intervals, the f(x) would give 0 for both the numbers in the interval. So, is it possible to have a abs. max/min on the same point?

No same point can't be both absolute min and max at the same time.

First you find the critical points $\displaystyle c = \{c_1, c_2, c_3, ...\}$ by letting $\displaystyle \frac{d(f(x))}{dx} = 0$ and solving for $\displaystyle x$

then take one critical point and say it's $\displaystyle c$ and do second derivative test. Second derivative test is:

if $\displaystyle \frac{d(f(c))}{dx^2} > 0$ then $\displaystyle c$ is absolute/local minimum.

or on the other hand if $\displaystyle \frac{d(f(c))}{dx^2} < 0$ then $\displaystyle c$ is absolute/local maximum.

Now if this second derivative test shows that you've multiple mins and/or maxes or you're asked to find local min/max in an interval ( $\displaystyle x = a$ to $\displaystyle x = b$ ) then plugin the values of critical points and/or both ends of interval into $\displaystyle f(x)$ and see which one is max and which one is min.

If second derivative of a function doesn't exist then use First derivative test which you can find by using Google.

Hope this helps.